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MATH2330: Analysis
MATH 2330: ANALYSIS
WADIM ZUDILIN & BRAILEY SIMS
Partly based on the lectures given in previous years at the University of Newcastle
y Dr Scott Sciffe
Contents
Topic 1 — Real numbers
Topic 2 — Sequences and convergence
Topic 3 — Limits of functions
Topic 4 — Series
Topic 5 — Differentiation
Topic 6 — Integration
Topic 7 — Extension to higher dimension
Date: 1st semester 2012.
MATH2330: ANALYSIS 1
We all have an intuitive understanding of what the real numbers are, or at least we
think we do. We also have a vague feeling about notions such as limits, continuity,
differentiation and integration. However the great mathematicians of a couple of
centuries ago discovered that vague intuition was not sufficient to prove “obvious
facts” about functions of a real variable. Indeed, sometimes these “obvious facts”
turned out to be not so obvious. . . and sometimes not even facts! It therefore
ecame necessary to be able to define these notions, which are central to the theory
of calculus, in a rigourous way. In doing so they uncovered many beautiful and
counterintuitive facts about the number system that we generally take for granted.
1. Real numbers
To warm up, let us start with the following
Exercise 1.1. Factorise (that is, expand into a product of non-trivial factors) the
following expressions:
(a) x4 − 1;
(b) e3x − 1;
(c) x2 − 2;
(d) x2 + 1; and
(e) x4 + 1.
Of course, the solution to each of the items might be not unique. For example, in
(a) one can do x4− 1 = (x− 1)(x3 + x2 + x+ 1) as well as x4− 1 = (x2− 1)(x2 + 1)
(the second way is in a certain sense better: one notifies that the factorisation may
e continued because x2 − 1 = (x− 1)(x+ 1)), while in (b) one can first substitute
y = ex and then factorise y3 − 1 (but what if you apply different substitutions, like
y = e3x/4 or even “simpler” y = e3x/2?). Expression in (c) can be done if you are
aware of i
ational (but still real !) numbers, while in (d) you are forced to go to the
world of complex numbers. As for (e), you can factorise remaining yourself in the
eal world (please do).
1.1. Field axioms. What is it we look for in a good number system? For a start
we would like to be able to add and multiply, and these operations should behave in
a “nice” fashion; for example, the order in which we add or multiply two numbers
should make no difference to the outcome. The natural numbers N do this much,
and N is a semigroup. But then we would like to be able to do subtractions, which
forces us to invent negative numbers and create the integers Z, and Z is a ring. But
if we want to be able to do division as well we have to go to the rationals Q. Recall
the rational numbers are
Q = {r : there exist p, q ∈ Z, q 6= 0 such that r = p/q}.
Remark. Note that there is a slight difference between the meanings of the words
ational and fraction. The term fraction refers to the representation of the numbe
2 WADIM ZUDILIN & BRAILEY SIMS
as a ratio of integers, whereas the term rational refers to any number which could
e represented in such a form. Thus both 1/2 and 0.5 are rational numbers, but
only 1/2 is a fraction (0.5 is a decimal).
The rational numbers form the strongest possible alge
aic structure, that of a
field. A set F is a field if it satisfies the following six properties with respect to
addition and multiplication and their inverses.
(1) Closure. If a, b ∈ F then a+ b ∈ F and ab ∈ F.
(2) Associativity. a+ (b+ c) = (a+ b) + c and (ab)c = a(bc) for all a, b, c ∈ F.
(3) Commutativity. a+ b = b+ a and ab = ba for all a, b ∈ F.
(4) Distributivity. a(b+ c) = ab+ ac for all a, b, c ∈ F.
(5) Identities. There exist unique elements of F, denoted 0 and 1, which are
the additive and multiplicative identities respectively; that is, a+ 0 = a and
1× a = a for all a ∈ F.
(6) Inverses. For any element a of F (except 0 in the multiplicative case) there
exist additive and multiplicative inverses denoted −a and a−1, respectively;
that is, a+ (−a) = 0 and a× a−1 = 1.
The rational numbers satisfy all these properties as do the real numbers and the
complex numbers. There are also finite fields — the integers modulo p where p is a
prime.
Exercise 1.2. Which of these axioms are not satisfied by N and Z?
From these axioms we can derive all the alge
aic laws of arithmetic of the real
numbers which we take for granted. For example,
Proposition 1.1 (Cancellation law). Suppose that for real numbers x, y and z,
x+ z = y + z. (1.1)
Then x = y.
Proof. Adding −z to each side of equation (1.1) yields
(x+ z) + (−z) = (y + z) + (−z),
and so
x+ (z + (−z)) = y + (z + (−z))
y the associative law for addition (Axiom 2). Then z+ (−z) = 0 (Axiom 6), hence
x+ 0 = y + 0 implying x = y by the property of 0 (Axiom 5). �
Remark. If two things are equal, and we do the same thing to both, then they remain
equal; so you may wonder why we do not just subtract from both sides in the theorem
above. The point is that subtraction is not an operation mentioned by the axioms.
The axioms only speak of the existence of an additive inverse. Therefore, in order to
subtract we have to add the inverse, which is how the proof proceeds. Try thinking
about the equivalent statement for multiplication. Is it true that if xz = yz then
x = y?
MATH2330: ANALYSIS 3
1.2. Order axioms. In addition to the above field axioms (alge
aic properties,
elated to equality) the rationals and reals also have a natural ordering (greate
than/less than) which satisfy the axioms below. These axioms give the reals an
analytic structure, which is a sense of “proximity” of elements. Once again these
axioms are not surprising and simply state specifically rules which we have used fo
years when dealing with the real numbers.
(7) Trichotomy. Given any a, b exactly one of the three statements a < b, a = b,
a > b is true.
(8) Transitivity. If a < b and b < c then a < c.
(9) Addition to inequality. If a < b then a+ c < b+ c.
(10) Multiplication of inequality. If a < b and c > 0 then ac < bc.
Once again, all the alge
aic properties of inequalities which we take for granted
can be derived from these axioms. For example,
Proposition 1.2. If x > 0 and y > 0 then x+ y > 0 (and similarly for negatives).
Proof. Since x > 0 (hence 0 < x), we have that 0 + y < x + y by Axiom 9. But
0 + y = y (Axiom 5) and y > 0 (given). Hence 0 < x+ y by Axiom 8. �
Proposition 1.3. If x > 0 then −x < 0.
Proof. Suppose −x > 0. Then Proposition 1.2 implies x + (−x) > 0. But since
−x is the additive inverse of x, we have x + (−x) = 0 (Axiom 6), and this would
contradict trichotomy (Axiom 7). So, −x cannot be positive.
Suppose −x = 0. Then x = x+ 0 = x+ (−x) = 0. But x cannot be both positive
and zero by trichotomy (Axiom 7). So −x cannot be zero.
Since −x is neither positive nor zero, it must be negative by trichotomy. �
Remark. In fact, this can be done much more easily. Start with x > 0 and use
Axiom 9 to add the inverse to both sides, and you are done. The above proof is
however a demonstration of the use of trichotomy, as well as a demonstration of how
to make something easy look much more impressive.
Exercise 1.3. Show that the following counterpart of Axiom 10 is true: if a < b and
c < 0 then ac > bc.
Exercise 1.4. Show that the additive and multiplicative identities (Axiom 5) satisfy
1 > 0.
1.3. Completeness. A set S is bounded above if there exists a number u such that
u ≥ s for every s ∈ S, and we say u is an upper bound for S. Similarly, S is bounded
elow if there exists a number l such that l ≤ s for every s ∈ S, and we say l is a
lower bound for S. We say a set is bounded if it is bounded above and below.
We say u is a least upper bound or supremum (sup) for S if u is an upper bound
for S, and no smaller upper bound exists. Similarly, l is a greatest lower bound o
infimum (inf) for S if l is a lower bound for S and no bigger lower bound exists.
The maximum of a set is the largest element of the set, if such an element exists;
that is, u is the maximum of the set S if u ∈ S and u ≥ s for all s ∈ S.
4 WADIM ZUDILIN & BRAILEY SIMS
The minimum of a set is the smallest element of the set, if such an element exists.
That is, l is the minimum of the set S if l ∈ S and l ≤ s for all s ∈ S.
It is absolutely central to the definitions of minimum and maximum that they
actually be elements of the set. Thus, for the set
S = {x ∈ R : 0 ≤ x < π}
we have inf(S) = min(S) = 0, and sup(S) = π, but S has no maximum because π is
not an element of S.
Exercise 1.5. Prove that if a set has a minimum (or a maximum), it must coincide
with the infimum (or supremum).
Exercise 1.6. Show that every non-empty finite set of real numbers is bounded, and
has a maximum and minimum element.
Proposition 1.4. The set R is not bounded above (or below).
Proof. Suppose that M is an upper bound for R. Then M + 1 belongs to R and
so M + 1 ≤ M . However, M + 1 > M (as follows from the order axioms and
Exercise 1.4). Supposing that R has an upper bound leads to a contradiction, hence
R is not bounded above. �
Proposition 1.5. Each set S has at most one supremum.
Proof. Let S be bounded above and let M1 and M2 be supremums of S. Since M1 is
an upper bound for S and M2 is a least upper bound, we have M2 ≤M1. Similarly,
since M2 is an upper bound for S and M1 is a least upper bound, M1 ≤ M2.
Therefore, M1 = M2. �
The method of showing x and y to be equal by showing that x ≤ y and y ≤ x is
used quite often. This method is an application of the trichotomy.
Exercise 1.7. Let S be a non-empty subset of R which has a lower bound. Show
that the set −S := {−x : x ∈ S} is bounded above and that − sup(−S) is a greatest
lower bound for S.
Solution. First, we show that − sup(−S) is a lower bound. Let x be in S. Then
−x ∈ −S and so −x ≤ sup(−S). Hence x ≥ − sup(−S).
Next we prove that − sup(−S) is greater than or equal to every lower bound for S.
To this end, let L be a lower bound. Then the derivation
−L ≥ −x for all x ∈ S =⇒ −L ≥ y for all y ∈ −S
shows that−L is an upper bound for−S, hence sup(−S) ≤ −L. Thus, − sup(−S) ≥
L. �
Our intuitive idea of the real numbers is motivated by the notion of a continuous
number line. Rational numbers are scattered all along the number line, but they are
not everything. In order to create the real numbers from the rationals we need to
find an axiom which will “plug up” the holes in the number line which the rationals
do not fill. This leads us towards the last axiom of the real numbers.
MATH2330: ANALYSIS 5
(11) Supremum or completeness axiom. Every non-empty set which is bounded
above has a least upper bound.
We now examine a few properties which are consequences of the supremum axiom.
Proposition 1.6 (Archimedean property). For any two numbers a, x ∈ R with
a > 0, there is an n ∈ N such that na > x.
Proof. Suppose that the statement is false for some a, x ∈ R. Then na ≤ x for every
n ∈ N and so x is an upper bound for the set A := {na : n ∈ N}.
The completeness axiom implies that A has a least upper bound. Put M =
sup(A). Then M − a < M and, consequently, M − a is not an upper bound for A.
Hence there is n ∈ N such that na > M − a. Adding a to each side yields, by
Axiom 9, that (n+ 1)a = na+ a > M , which contradicts M being an upper bound
ecause (n+ 1)a ∈ A. Therefore, the original supposition is false. �
Remark. Strictly speaking we do not need the supremum axiom to prove this, since
Q also has this property; however it is easier to prove via the supremum axiom.
The Archimedean property allows us to say that between any two distinct real
numbers there is another real number. The way this is commonly used is to say that
for any positive number, no matter how small, there is a smaller positive number.
In order to see what is wrong with the field of rationals, recall the following fact
(cf. Exercise 1.1(c)).
Proposition 1.7 (I
ationality of
√
2). There is no rational number whose square
is 2.
Proof. Suppose there is such a number r, so r = p/q where p, q ∈ Z and we may
assume that p, q have no common factors. Then r2 = p2/q2 = 2, so p2 = 2q2. Thus
p2 is an even number, and therefore p must be even (since the square of an odd
number is always odd). That is, p = 2n for some n ∈ Z. Then (2n)2 = 2q2 implying
2n2 = q2, and by similar reasoning we conclude that q is an even number. But if p
and q are both even, they have a common factor of 2, contradicting our assumption
of no common factors. We therefore conclude that no such rational number r can
exist. �
To understand how drastically i
ational numbers rocked the Greek world we need
to understand how they viewed the relationship between geometry and numbers. It
was long held that any two line segments were commensurate; that is, it would be
possible to find a third line segment such that the original two were whole numbe
multiples of the third. This is equivalent to saying the ratio of the lengths of the
original two line segments is rational. The existence of a line segment of length
√
2
(as the hypotenuse of the {1, 1,
√
2} right triangle) destroyed this belief. You have
doubtless seen the proof of the i
ationality of
√
2 before, but as the great numbe
theorist Hardy once said, it “is as fresh and significant as when it was discovered —
two thousand years have not written a wrinkle on [it].”
Let us see how the supremum axiom guarantees us the existence of
√
2 in R.
6 WADIM ZUDILIN & BRAILEY SIMS
Lemma 1.1. Suppose that w2 > 2 and w > 0. Then w is an upper bound fo
S = {x ∈ Q : x2 < 2}.
Proof. Suppose that y > w. Then y2 > w2 > 2 and so y /∈ S. Hence x ≤ w
whenever x ∈ S. �
Proposition 1.8. There exists a real number r such that r2 = 2.
Proof. Let S = {x ∈ Q : x2 < 2}. Then 22 > 2 and so 2 is an upper bound for S by
Lemma 1.1. By the supremum axiom S has a least upper bound, put r = supS.
First check that r2 6< 2. If r2 were less than 2, we could find w > r such that
w2 < 2. Then r would not be an upper bound.
Secondly, we must have r2 6> 2. Indeed, if r2 were bigger than 2, we could find
w < r such that w2 > 2. Lemma 1.1 shows that w is an upper bound and so r would
not be the least upper bound for S.
It follows from the above that r2 = 2 by trichotomy. In other words, r is the
square root of 2. �
The above argument shows that any ordered field which satisfies the completeness
axiom contains
√
2. Since
√
2 is not rational, it follows that Q does not satisfy
Axiom 11.
Proposition 1.9. The ordered field (Q,+,×, <) is not complete, that is, there is a
subset of Q which is bounded above but does not have a supremum.
Theorem 1.1. The field (R,+,×, <) of real numbers is the only complete ordered
field.
We shall not prove this! However, to clarify what the statement of the theorem
means, if we have a complete ordered field F, then there is a unique bijection θ : F→
R which preserves all the field and order structures. Thus, if 0F and 0R are the
espective zeros in F and R, then θ(0F) = 0R. Also, θ(x+ y) = θ(x) + θ(y), θ(xy) =
θ(x)θ(y) and x < y if and only if θ(x) < θ(y).
There are a number of different conditions which are equivalent to the supremum
axiom. We shall see some of them when we study sequences and functions.
MATH2330: ANALYSIS 7
1.4. Absolute value. The absolute value of a real number x is
|x| =
{
x if x ≥ 0,
−x if x < 0.
The absolute value function is such an important function because of its geometrical
interpretation. The absolute value of a number is its distance from the origin. Thus
we should interpret |x| as “the distance of x from 0”. In a similar way |x − a|
should be thought of as “the distance from x to a”. Thinking this way often makes
problems easier.
It is difficult to understand the term “triangle inequality” when dealing with the
eal numbers, because it is impossible to draw a proper triangle in the number line.
Perhaps it is best to describe the triangle inequality in R2 (or even in R3), where
triangles exist, and then relate it to R as degenerate case. Consider the triangle
in R2 with sides a, b and a + b. As we know from geometry,
‖a + b‖ ≤ ‖a‖+ ‖b‖
and equality occurs if and only if the two vectors a and b are parallel (and pointing
the same way), or if one of them is 0. For the real case imagine flattening the triangle
until all three vertices are colinear (can you still call it a triangle?) and view these
points as lying on the number line. Now in R the vectors are just numbers, and
there are only two directions in which a vector can point — the positive or negative
directions. It follows that on the real line the norm of a vector is just the absolute
value of a number. Thought of this way, our triangle inequality in R2 squashes down
to
|a+ b| ≤ |a|+ |b|
on R, and equality occurs if and only if a and b have the same sign, or if one (o
oth) of them is zero.
Now let’s prove it formally.
Proposition 1.10 (Triangle inequality). For every a, b ∈ R, |a+ b| ≤ |a|+ |b|.
Proof. If either a or b are zero, or if a and b have the same sign, then |a+b| = |a|+|b|.
Otherwise one of a or b is positive and the other is negative. Without loss of
generality we may assume a > 0 and b < 0. We now consider two possibilities.
Case 1: |a| > |b|. In this case a+ b > 0 so
|a+ b| = a+ b ≤ a = |a| ≤ |a|+ |b|.
Case 2: |a| < |b|. In this case a+ b < 0 so
|a+ b| = −(a+ b) ≤ −b = |b| ≤ |a|+ |b|.
Thus, we have demonstrated that the inequality holds in all circumstances. �
Exercise 1.8. Show that |a+ b+ c| ≤ |a|+ |b|+ |c|.
Solution. Indeed, applying twice the triangle inequality we get
|(a+ b) + c| ≤ |a+ b|+ |c| ≤ |a|+ |b|+ |c|.
�
8 WADIM ZUDILIN & BRAILEY SIMS
We say that a set N is a neighbourhood of x ∈ R if there is an open interval I
containing x and contained in N . If this is the case, then for some small positive
number ε we have (x − ε, x + ε) ⊆ N . Then we sometimes describe N as an ε-
neighbourhood of x, because it contains all things around x within a distance ε.1
Given subsets A and B of the reals, we say A is dense in B if in every neighbour-
hood of every point of B there are elements of A; that is,
∀b ∈ B and ε > 0, A ∩ {x : |b− x| < ε} 6= ∅.
Proposition 1.11. Both the rationals and i
ationals are dense in the reals.
Proof. Let r be real. Let rn be the decimal expansion of r up to n decimal places.
Then each rn is a rational number, and r is the supremum of all the rn. It follows
that for any ε > 0 eventually we have r − ε < rn ≤ r, otherwise r would not be the
supremum. Since any neighbourhood must contain (r − ε, r + ε) for some ε > 0 we
see that the rationals get inside any neighbourhood of r. So the rationals are dense
in the reals.
Let r be real. If r is i
ational then there is nothing to prove, since r is auto-
matically in any neighbourhood of itself. If r is rational, let rn = r +
√
2/n, so rn
is i
ational. Now for any ε > 0 eventually (for big enough values of n) we have√
2/n < ε, so r < rn < r + ε. Since any neighbourhood must contain (r − ε, r + ε)
for some ε > 0 we see that the i
ationals get inside any neighbourhood of r. Thus,
the i
ationals are dense in the reals. �
2. Sequences and convergence
The intuitive notion of a sequence is an unending list of numbers. Notice that
implicit in this notion is that the ordering of the numbers is important, which is
what distinguishes a sequence of numbers from a mere (countable) set of numbers.
More formally, a sequence is a function whose domain is the natural numbers; that
is, x : N→ R is a sequence.
Rather than the usual function notation, where the argument of the function is
appended in
ackets, for sequences it is customary to use a subscript. Thus we
indicate x is a function of n by writing xn rather than x(n). Each individual entry
in a sequence is called a term.
Remark. A sequence, like any function, need not be given by an explicit formula.
Sometimes a sequence is defined by some iterative process. You have seen this
already in Newton’s method for approximating roots of an equation. Recall that to
find an approximate solution to the equation f(x) = 0 you have a guess, x1, and
then iterate the equation
xn+1 = xn −
f(xn)
f ′(xn)
.
1Get used to the symbol ε (epsilon) because it will soon drive you insane. Whenever a mathe-
matician wants to describe a small positive number they call it ε, unless of course they have already
used ε, in which case they call it δ (delta).
MATH2330: ANALYSIS 9
This produces a sequence of numbers which hopefully get closer and closer to the
true solution of the equation. But in general we cannot assume that a sequence has
any particular pattern relating its terms. It could be essential random, but it is still
a sequence.
2.1. Convergence of a sequence. Consider the sequence xn = 1/n. Now as the
value of n increases the value of 1/n decreases. In fact with the intuitive notion of
limits you have gained from calculus you should not be surprised if we claimed
lim
n→∞
1
n
= 0.
But now we have to say precisely what this means. Clearly it does not mean 1/n
ever equals zero, because it never does. So it must mean that 1/n is nearly equal to
zero. But how nearly? Well, eventually we want it to be a
itrarily close. How do
we say that mathematically?
A
itrarily close? We have to pick any measure of closeness, say ε > 0. So we want
to make |1/n− 0| < ε.
Eventually? Then, using the value of ε we have to find a term of the sequence, say
the N -th term, beyond which every term has the desired degree of closeness. In ou
particular case we just need to choose our N = 1/ε because if n > N = 1/ε then∣∣∣∣ 1n − 0
∣∣∣∣ = 1n < 1N = 11/ε = ε.
This example motivates our definition.
Formally, a sequence xn tends to a limit L, or converges to L, o
lim
n→∞
xn = L
if, given any ε > 0, there exists N (depending on ε) such that every term past the
N -th term is within ε distance of L. In symbols this is written
∀ε > 0 ∃N(ε) such that n > N =⇒ |xn − L| < ε.
It is possible to use the definition to verify the limit of a sequence, but in ou
first examples you need to be able to guess the limit beforehand, and the method is
tedious except on the simplest of cases.
Remark. The idea of limit is implicit in the notion of decimal expansion (cf. the
proof of Proposition 1.11). Check that, if xn is π to n decimal places (x1 = 3.1,
x2 = 3.14, x3 = 3.141, . . . ), then xn tends to π.
Proposition 2.1 (Uniqueness of limit). Suppose that xn → L and xn → M . Then
L = M .
Proof. Suppose that L 6= M . Then |L −M | > 0. We use the definition of limit
taking ε = |L −M |/2 > 0. Since xn → L, there exists N1 such that |xn − L| < ε
whenever n ≥ N1; similarly, there is N2 such that |xn −M | < ε whenever n ≥ N2.
Using the triangle inequality we have for n ≥ max{N1, N2},
|L−M | = |(xn −M)− (xn − L)| ≤ |xn −M |+ |xn − L| < 2ε = |L−M |.
This contradiction implies that the supposition that L 6= M is false. �
10 WADIM ZUDILIN & BRAILEY SIMS
Exercise 2.1. Find the limit of xn = (n+ 2)/(2n+ 1).
Solution. We first need to guess the limit: n = 1 gives xn = 1, n = 10 gives xn =
12
21
= 0.57 . . . , n = 100 gives xn = 0.507 . . . and n = 1000 gives xn = 0.5007 . . . . It
would appear that we can take L = 1/2 as the limit. Let us verify it:
|xn − L| =
∣∣∣∣ n+ 22n+ 1 − 12
∣∣∣∣ = ∣∣∣∣(2n+ 4)− (2n+ 1)4n+ 2
∣∣∣∣ = ∣∣∣∣ 34n+ 2
∣∣∣∣ = 34n+ 2 .
Now we need to be able to make this less than any ε. So
3
4n+ 2
ε ⇐⇒ 3
ε
4n+ 2 ⇐⇒ 1
4
(
3
ε
− 2
)
n.
That is, given any ε > 0 there exists N = 1
4
(3/ε− 2) such that if n > N then∣∣∣∣ n+ 22n+ 1 − 12
∣∣∣∣ < ε.
So the limit is a half. �
What happens if we make a bad guess? Suppose in the above example we guessed
that the limit was 2. Then
|xn − L| =
∣∣∣∣ n+ 22n+ 1 − 2
∣∣∣∣ = ∣∣∣∣ −3n2n+ 1
∣∣∣∣ = 3n2n+ 1 .
Now we need to be able to make this less than any ε. So
3n
2n+ 1
ε ⇐⇒ 3
ε
2+
1
n
⇐⇒ 3
ε
−2 < 1
n
⇐⇒ ε
3− 2ε
n for small enough ε.
Note this gives an upper bound on n which will work, not a lower bound. Hence we
have found a term of the sequence beyond which the required inequality does not
hold. Thus, 2 is not the limit of the sequence.
Exercise 2.2. Find the limit of xn = 1 + (−1/2)n.
Solution. Computing a few first terms, it looks like the limit is L = 1. To prove it,
given ε > 0 we need
|xn − L| < ε ⇐⇒
∣∣∣∣(1 + (−12
)n)
− 1
∣∣∣∣ < ε ⇐⇒ 12n < ε ⇐⇒ 2n > 1ε
⇐⇒ n ln(2) > ln(1/ε) ⇐⇒ n > − ln(ε)
ln(2)
.
Note that it is legitimate to take logarithms of both sides of an inequality if they
are positive, because ln(x) is a monotone increasing function and therefore preserves
the inequality.
This calculation tells us exactly how large we need to make N in order to achieve
the nominated degree of closeness. Thus,
∀ε > 0 ∃N(ε) = − ln(ε)
ln(2)
such that n > N =⇒
∣∣∣∣(1 + (−12
)n)
− 1
∣∣∣∣ < ε
and we are done. �
MATH2330: ANALYSIS 11
Exercise 2.3. Find the limit of xn = n/2
n.
Solution. After some trials we guess that the limit is L = 0. The required inequality
|xn − L| < ε is equivalent to n/2n < ε— it is difficult to solve it explicitly because
the expression in n on the left is “too transcendental”. But note that the binomial
theorem implies
2n = (1 + 1)n = 1 + n · 1 + n(n− 1)
2
· 1 + · · ·+ n(n− 1)
2
· 1 + n · 1 + 1
2 + 2n+ n(n− 1) = n2 + n+ 2 > n2
for all n ≥ 5, so that n/2n < n/n2 = 1/n for those n. Therefore, if we assume
1/n < ε ⇐⇒ n > 1/ε, equivalently, if we take N(ε) = max{5, 1/ε}, we do get
|xn − L| = n/2n < ε. �
Finally we try something a bit uglier.
Exercise 2.4. Show that
n+ 37
n5 − 2n2
→ 0 as n→∞.
Solution. Fix ε > 0. We want N such that
n > N =⇒
∣∣∣∣ n+ 37n5 − 2n2 − 0
∣∣∣∣ < ε. (2.1)
Since n+ 37 is always positive, we have∣∣∣∣ n+ 37n5 − 2n2 − 0
∣∣∣∣ = n+ 37n5 − 2n2
provided that the bottom line is positive. This holds if n > 3
√
2.
The inequality (n+ 37)/(n5− 2n2) < ε is difficult to solve because the expression
in n on the left is complicated. Fortunately, we do not need to find all solutions
ut only values of n large enough that are guaranteed to solve the inequality (cf.
with our previous example). We can do this by replacing (n + 37)/(n5 − 2n2) by
something bigger but simpler. We can see in this example that the reason the limit
is 0 is that n5 grows much faster than n. The idea is to make use of that fact and
make precise estimates.
First, we have for all n ∈ N that n + 37 ≤ n + 37n = 38n. Secondly, n5 − 2n2 ≥
n5 − n5/2 provided that 2n2 ≤ n5/2; that is, n3 ≥ 4, which is satisfied by n ≥ 3
√
4.
Hence
n+ 37
n5 − 2n2
≤ 38n
n5 − n5/2
=
76
n4
provided that n ≥ 3
√
4 (since then n > 3
√
2 is also true). This inequality implies
(2.1) with N = max{ 3
√
4, 4
√
76/ε}. �
Exercise 2.5. Find the limit of xn = (2n+ 2
n)/(3n+ 3n).
Exercise 2.6. Prove that for p > 0,
lim
n→∞
1
np
= 0.
12 WADIM ZUDILIN & BRAILEY SIMS
Exercise 2.7. Compute
lim
n→∞
(
1
n2
+
2
n2
+ · · ·+ n− 1
n2
)
.
Hint. First show by induction on n that
1 + 2 + · · ·+ (n− 1) = n(n− 1)
2
and then apply the standard machinery. �
2.2. Alge
a of limits. Clearly, we do not want to have to use the definition all
the time! We therefore prove the following theorem, which allows us to
eak down
the limits of difficult expressions into combinations of simpler expressions for which
the limits are known.
Theorem 2.1 (Alge
a of limits theorem — ALT). If an and bn are sequences which
converge to a and b, respectively, then
(i) lim
n→∞
(an + bn) = a+ b;
(ii) lim
n→∞
(anbn) = ab;
(iii) lim
n→∞
(λan) = λa for any constant λ ∈ R;
(iv) lim
n→∞
an
n
=
a
provided b 6= 0; and
(v) lim
n→∞
(an)
p = ap for any constant p 6= 0.
Proof. We shall only prove the first two cases.
(i) Given any ε > 0, we take an even smaller number ε′ = ε/2. Since an → a,
there exists an Na such that for every n > Na we have |an− a| < ε′; similarly, there
exists an Nb such that n > Nb implies |bn − b| < ε′. Then conside
|(an + bn)− (a+ b)| ≤ |an − a|+ |bn − b|
y the triangle inequality. Now if n > max{Na, Nb} then we have
|(an + bn)− (a+ b)| ≤ |an − a|+ |bn − b| < ε′ + ε′ = ε.
(ii) Given any ε > 0, we calculate an even smaller number ε′ = ε/(|a| + |b| + 1),
and if this is still bigger than 1, take ε′ = 1. Since an and bn tend to a and b,
espectively, we can find numbers Na and Nb as in (i) above. Using the triangle
inequality,
|anbn − ab| = |anbn − anb+ anb− ab| ≤ |anbn − anb|+ |anb− ab|
= |an| |bn − b|+ |b| |an − a|.
Now for n > max{Na, Nb} we have
|anbn − ab| ≤ |an| |bn − b|+ |b| |an − a| ≤ (|a|+ ε′)ε′ + |b|ε′ = ε′(|a|+ |b|+ ε′)
≤ ε′(|a|+ |b|+ 1) ≤ ε,
the desired estimate. �
MATH2330: ANALYSIS 13
Judicious use of Theorem 2.1, along with known limits (like 1→ 1 and 1/n→ 0),
allows us to do a wide variety of problems quickly and easily. In the following
example all the steps are laid out labouriously.
Exercise 2.8. Compute
lim
n→∞
2n2 + 3n+ 5
n3 + n2 + 1
.
Solution. Dividing the numerator and denominator by the fastest growing function
n3 we obtain
lim
n→∞
2n2 + 3n+ 5
n3 + n2 + 1
= lim
n→∞
2
n
+ 3
n2
+ 5
n3
1 + 1
n
+ 1
n3
=
limn→∞
(
2
n
+ 3
n2
+ 5
n3
)
limn→∞
(
1 + 1
n
+ 1
n3
) (by ALT (iv))
=
limn→∞
2
n
+ limn→∞
3
n2
+ limn→∞
5
n3
limn→∞ 1 + limn→∞
1
n
+ limn→∞
1
n3
(by ALT (i))
=
2 limn→∞
1
n
+ 3 limn→∞
1
n2
+ 5 limn→∞
1
n3
1 + limn→∞
1
n
+ limn→∞
1
n3
(by ALT (iii))
=
2 · 0 + 3 · 0 + 5 · 0
1 + 0 + 0
(by ALT (v))
=
0
1
= 0.
Of course, we do not really want to have lay out every tiny step. In our example
the following is quite sufficient,
lim
n→∞
2n2 + 3n+ 5
n3 + n2 + 1
= lim
n→∞
2
n
+ 3
n2
+ 5
n3
1 + 1
n
+ 1
n3
=
0
1
= 0
y Theorem 2.1. �
Exercise 2.9. Compute
lim
n→∞
(
1
2
+
3
22
+
5
23
+ · · ·+ 2n− 1
2n
)
.
Solution. Note that
xn −
1
2
xn =
(
1
2
+
3
22
+
5
23
+ · · ·+ 2n− 1
2n
)
−
(
1
22
+
3
23
+ · · ·+ 2n− 3
2n
+
2n− 1
2n+1
)
=
1
2
+
(
2
22
+
2
23
+ · · ·+ 2
2n
)
− 2n− 1
2n+1
=
1
2
+
(
1− 1
2n−1
)
− 2n− 1
2n+1
=
3
2
− 2n+ 3
2n+1
=
3
2
− n
2n
− 3
2
· 1
2n
→ 3
2
− 0− 3
2
· 0
y Theorem 2.1 and Exercise 2.3. Thus xn → 3 as n→∞. �
14 WADIM ZUDILIN & BRAILEY SIMS
This is another tool for determining the limit of a complicated sequence by com-
paring it with known limits of simpler sequences. We used it implicitly in Exer-
cises 2.3 and 2.4.
Theorem 2.2 (Squeeze theorem). (i) If an, bn and cn are sequences such that
an ≤ bn ≤ cn and an → L and cn → L, then bn → L.
(ii) If an and bn are sequences such that an ≤ bn and an → ∞, then bn → ∞.
Similarly, if bn → −∞ then so does an.
Proof. (i) Given ε > 0, there exists an N1 such that for n > N1 we have |an−L| < ε.
More particularly, L−ε < an. Similarly, there exists an N2 such that n > N2 implies
that cn < L+ ε. So for n > max{N1, N2} we get
L− ε < an ≤ bn ≤ cn < L+ ε.
(ii) First, we need a definition of an →∞! This means that now matter how big
(M) one asks us to make the sequence, we can find a term of the sequence (N -th
term) beyond which (n > N) every term is bigger than M . Symbolically,
∀M ∃N = N(M) such that n > N =⇒ an > M.
Now for every M there exists N such that n > N implies an > M , and since bn ≥ an
the same choice of N will do to show bn →∞. �
It does not really stretch the imagination to believe this theorem, but how is it
useful?
Note that
0
1
n
√
n
≤ 1
n
,
and the sequences 0 and 1/n both converge to 0. Thus, by the squeeze theorem so
does the sequence 1/(n
√
n).
For an = (−1)n/n we have −1/n ≤ an ≤ 1/n, so (−1)n/n converges to 0 by the
squeeze theorem.
For an = n
3/(n2 − 1) we have
n3
n2 − 1
n3
n2
= n→∞.
Thus, an diverges to ∞.
MATH2330: ANALYSIS 15
The next result traditionally accompanies the squeeze theorem and looks useless,
ut please check how remarkably it works in Exercise 2.15 (coming soon).
Proposition 2.2. Suppose that an → a and bn → b as n→∞, and eventually (fo
all n > N for some N), an ≤ bn. Then a ≤ b.
Proof. As in the proof of Theorem 2.2, given ε > 0, there exist N1 and N2 such
that for n > N1 we have a − ε < an and for n > N2, bn < b + ε. Then fo
n > max{N,N1, N2} we have a − ε < an ≤ bn < b + ε. Assuming that a > b and
taking ε = (a−b)/2 in the last estimate we get a−b < 2ε = a−b, a contradiction. �
Exercise 2.10. Show that
n+ 42 sin 6n
n3 − 3n
→ 0 as n→∞.
Solution. Start by dividing through by n3,
n+ 42 sin 6n
n3 − 3n
=
1/n2 + (42 sin 6n)/n3
1− 3/n2
.
We already know that 1/n2 → 0 and 1 → 1 as n → ∞. What about the limit
of bn = (42 sin 6n)/n
3? It cannot be obviously evaluated with the alge
a of limits
ecause 42 sin 6n does not converge as n→∞. We use the squeeze theorem instead.
Since −1 ≤ sinn ≤ 1 for all n, we have
−42
n3
≤ 42 sin 6n
n3
≤ 42
n3
for all n. The sequences an = −42/n3 and cn = 42/n3 both converge to 0 as n→∞,
so does the sequence bn. �
Exercise 2.11. Find the limit of xn =
√
n+ 1−
√
n.
Hint. Write
√
n+ 1−
√
n =
1√
n+ 1 +
√
n
1
2
√
n
and use Exercise 2.6 and Theorems 2.1, 2.2. �
Exercise 2.12. Show the inequality
1
2
· 3
4
· · · 2n− 1
2n
1√
2n+ 1
and deduce from it that
lim
n→∞
1
2
· 3
4
· · · 2n− 1
2n
= 0.
Hint. Use mathematical induction. �
Proposition 2.3 (Bernoulli inequality). If x > −1, then (1 + x)n ≥ 1 + nx for all
integers n > 0. Moreover, the equality happens if only n = 1 or x = 0.
16 WADIM ZUDILIN & BRAILEY SIMS
Proof. If n = 1 we get the equality (1 + x)1 = 1 + x which also means the above
inequality. Suppose now that n > 1 and we have already shown that
(1 + x)n−1 ≥ 1 + (n− 1)x
for the preceding exponent. Multiplying the both sides of the latter by 1 + x > 0
esults in
(1 + x)n ≥ (1 + (n− 1)x)(1 + x) = 1 + nx+ (n− 1)x2 ≥ 1 + nx,
with the equality possible if only x = 0 (recall that n > 1). Thus, the mathematical
induction principle justifies the required inequality for all n ≥ 1. �
Exercise 2.13. Prove that for a > 0,
lim
n→∞
n
√
a = 1.
Solution. The case a = 1 is clear. Let a > 1. Then n
√
a > 1 for all n and the
Bernoulli inequality implies
a =
(
1 + ( n
√
a− 1)
)n
1 + n( n
√
a− 1) > n( n
√
a− 1),
hence 0 < n
√
a − 1 < a/n and the desired claim follows from the squeeze theorem.
If 0 < a < 1,
lim
n→∞
n
√
a = lim
n→∞
1
n
√
1/a
=
1
lim
n→∞
n
√
1/a
= 1
y the ALT. �
Exercise 2.14. Prove the following limits:
(a) lim
n→∞
nk
an
= 0 (a > 1), (b) lim
n→∞
loga n
n
= 0 (a > 1).
2.3. Monotone convergence. Euler’s constants. The two preceding theorems,
Theorems 2.1 and 2.2, assume the existence of the limits of some known sequences,
whose limits can be guessed and verified from the definition. Now we have to tackle
the problem of proving that a sequence has a limit, even if it is not obvious what
that limit is. As you will see, the supremum axiom has a major role to play here, as
it must, since it is the only mechanism by which we know most of the real numbers
exist. Before embarking on the proof of our next theorem we shall need the following
definitions.
A sequence an is increasing (strictly increasing) if an+1 ≥ an (an+1 > an) fo
every natural number n; similarly for decreasing (strictly decreasing). A sequence
is monotone if it is either increasing or decreasing. A sequence satisfies any of the
above properties eventually if there exists N such that the properties hold for every
natural number n > N .
Theorem 2.3 (Monotone convergence theorem). If an is (eventually) monotone
increasing (decreasing) and bounded above (respectively, below), then an converges
to some limit L.
MATH2330: ANALYSIS 17
Proof. If an is bounded above, it has a least upper bound, by the supremum axiom.
Call this least upper bound L. We aim to show L is the limit of the sequence. Fo
any ε > 0 there exists N such that L − ε < aN ≤ L, otherwise L would not be a
least upper bound. But then since an is increasing we have that for every n > N ,
an > aN . Thus L− ε < aN ≤ an ≤ L, and the limit is L.
The decreasing case follows similarly, and if the sequence is eventually mono-
tone simply apply the argument to that portion of the sequence beyond which the
monotonicity starts. �
Consider the sequence an = 1/2
n. Now since an+1/an = 1/2, we see that the
sequence is decreasing. Also, since all terms are positive, it is bounded below by
zero. Thus by Theorem 2.3 it has a limit. We do not know what this limit is! Give
it a name, L say. Then since an+1 =
1
2
an, if we apply the ALT (Theorem 2.1) to the
oth sides(!) we see that L = 1
2
L. Thus L = 0.
In the example above we could compute the limit of an without reference to the
monotone convergence theorem. Our next example is more fundamental and makes
heavy use of Theorem 2.3. It also explains on how Euler’s constant
e = 2.71828182845904523536028747 . . . , (2.2)
the base of natural logarithm, is mathematically defined.
Proposition 2.4 (Euler’s constant e). The sequences
xn =
(
1 +
1
n
)n
and yn =
(
1 +
1
n
)n+1
tend to the same limit, which we call e:
e = lim
n→∞
(
1 +
1
n
)n
= lim
n→∞
(
1 +
1
n
)n+1
.
Proof. Let us show that the sequence xn is strictly increasing and bounded above,
while the sequence yn is strictly decreasing and bounded below. To show the mono-
tonicity we use the Bernoulli inequality (Proposition 2.3),
xn+1
xn
=
(
1 +
1
n+ 1
)n+1(
1 +
1
n
)
(
1 +
1
n
)n+1 = (1− 1(n+ 1)2
)n+1
n+ 1
n
(
1− 1
n+ 1
)
n+ 1
n
= 1,
18 WADIM ZUDILIN & BRAILEY SIMS
yn
yn−1
=
(
1 +
1
n
)n(
1 +
1
n
)
(
1 +
1
n− 1
)n = 1(
1 +
1
n2 − 1
)n · n+ 1
n
1
1 +
n
n2 − 1
· n+ 1
n
=
n3 + n2 − n− 1
n3 + n2 − n
1.
Since also xn < yn for all n, the monotonicity implies that xn < y1 = 4 and
yn > x1 = 2 for all n. We conclude from Theorem 2.3 that the limits of xn and yn
exist. Furthermore,
0 < yn − xn =
1
n
(
1 +
1
n
)n
y1
n
→ 0 as n→∞,
hence the limits in fact coincide. It is this joint limit which is called e, after Euler. �
Corollary. For n = 1, 2, . . . ,(
1 +
1
n
)n
e
(
1 +
1
n
)n+1
. (2.3)
It is impractical to use the definition of e from Proposition 2.4 for calculating the
constant: for instance, x1000 = 2.7169 . . . — nothing but 3 co
ect decimal places.
Recall a similar story with geometric definition of π— one needs to inscribe a regula
polygon with an enormous number of vertices just to get 3 digits in π = 3.14 . . . .
With some effort, we can derive a more practical expression for e.
Exercise 2.15. Prove that
lim
n→∞
(
1 +
1
1!
+
1
2!
+
1
3!
+ · · ·+ 1
n!
)
= e.
Solution. The only definition of e (on the right-hand side) which we can use in ou
derivation is the one from Proposition 2.4.
By Newton’s binomial formula,(
1 +
1
n
)n
= 1 + n · 1
n
+
n(n− 1)
2
· 1
n2
+ · · ·+ n(n− 1) · · · (n− k + 1)
k!
· 1
nk
+ · · ·
+
n(n− 1) · · · 2 · 1
n!
· 1
nn
= 2 +
1
2!
(
1− 1
n
)
+ · · ·+ 1
k!
(
1− 1
n
)
· · ·
(
1− k − 1
n
)
+ · · ·
+
1
n!
(
1− 1
n
)
· · ·
(
1− n− 1
n
)
2 +
1
2!
+ · · ·+ 1
n!
=: zn.
MATH2330: ANALYSIS 19
Moreover, if we fix a positive integer k, then for all n > k we have(
1 +
1
n
)n
2 +
1
2!
(
1− 1
n
)
+ · · ·+ 1
k!
(
1− 1
n
)
· · ·
(
1− k − 1
n
)
(we simply truncate after the k-th summand the tail of the sum from the above
derivation). Since both sides of the latter inequality have finite limits as n → ∞,
passing to them with the help of Proposition 2.2 we obtain
e ≥ 2 + 1
2!
+ · · ·+ 1
k!
= zk.
To summarise our findings, (
1 +
1
k
)k
zk ≤ e,
and application of the squeeze theorem shows that zk → e as k → ∞, as required.
�
Note that already the 26-th term
z26 = 1 +
1
1!
+
1
2!
+
1
3!
+ · · ·+ 1
26!
gives e with accuracy recorded in (2.2). The result of Exercise 2.15 can be used to
show that e is i
ational; more is known — this number is transcendental (not a root
of polynomial with rational coefficients). A much more mysterious number, at least
from the (ir)rational point of view, is another Euler’s constant,
γ = 0.57721566490153286060651209008240243104215933593992 . . . .
Proposition 2.5 (Euler’s constant γ). The sequence
xn = 1 +
1
2
+
1
3
+ · · ·+ 1
n
− lnn
converges to a finite limit. The limit is called γ.
Proof. Taking the logarithm of all sides in (2.3) we see that
1
n+ 1
ln
(
1 +
1
n
)
1
n
. (2.4)
Furthermore, since
xn+1 − xn =
1
n+ 1
− ln(n+ 1) + lnn = 1
n+ 1
− ln
(
1 +
1
n
)
0
y (2.4), the sequence xn is strictly decreasing. It is also bounded below:
xn = 1 +
1
2
+
1
3
+ · · ·+ 1
n
− lnn
ln(1 + 1) + ln
(
1 +
1
2
)
+ · · ·+ ln
(
1 +
1
n
)
− lnn
= ln
(
2 · 3
2
· 4
3
· · · n+ 1
n
· 1
n
)
= ln
n+ 1
n
1
n+ 1
0.
20 WADIM ZUDILIN & BRAILEY SIMS
Thus, the limit of xn exists by Theorem 2.3 (and is positive by Proposition 2.2). �
Exercise 2.16. Let the sequence xn be defined by the following formula:
x1 > 0, xn+1 =
1
2
(
xn +
1
xn
)
for n = 1, 2, . . . .
Prove that
lim
n→∞
xn = 1.
Solution. Since x1 > 0 and xn + 1/xn ≥ 2 (the latter follows from the relation
etween arithmetic and geometric means:
√
ab ≤ (a + b)/2), the sequence xn is
ounded below by 1. Hence 1/xn ≤ 1 ≤ xn and from the inequality
xn+1 =
1
2
(
xn +
1
xn
)
≤ xn,
which is valid for n ≥ 2, it follows that the sequence is decreasing. By Theorem 2.3
it tends to a limit, say x; Proposition 2.2 ensures that x ≥ 1. Passing to the limit
n→∞ in
xn+1 =
1
2
(
xn +
1
xn
)
(Theorem 2.1) we get x = 1
2
(x + 1/x). Solving it we find that x = 1 (anothe
possibility x = −1 is excluded by x ≥ 1). �
Exercise 2.17. Show that the sequence
x1 =
√
2, x2 =
√
2 +
√
2, . . . , xn =
√
2 +
√
2 + · · ·+
√
2︸ ︷︷ ︸
n radicals
, . . .
converges and compute its limit.
Hint. Show the recu
ence xn+1 =
√
2 + xn and use induction to prove that xn < 2
and that the xn is increasing. �
Exercise 2.18. Using Theorem 2.3 show that the sequence
xn =
(
1 +
1
2
)(
1 +
1
4
)
· · ·
(
1 +
1
2n
)
converges.
Exercise 2.19 (Arithmetic-geometric mean). Prove that the sequences xn and yn
constructed by the relations
x1 = a > 0, y1 = b > 0, xn+1 =
√
xnyn, yn+1 =
xn + yn
2
,
converge to a joint limit,
µ(a, b) = lim
n→∞
xn = lim
n→∞
yn.
(The limit is called the arithmetic-geometric mean of a and b.)
MATH2330: ANALYSIS 21
2.4. Cauchy sequences. Not every sequence is monotone, and so we need more
theorems to prove convergence results for sequences.
A subsequence of the sequence xn is any sequence obtained by deleting some
(finitely or infinitely many) of the terms of xn. We write xnk for the k-th term of
the subsequence. Thus n1 < n2 < n3 < · · · ; note that nk ≥ k for all k.
There are different ways to describe subsequences of a sequence. For example, the
sequences 1/2k, 1/k2 and 1/(2k + 1) are all subsequences of 1/n: we take nk = 2
k,
nk = k
2 and nk = 2k + 1, respectively. Another example: if xn is an a
itrary
sequence with infinitely many positive terms, then discarding all the non-positive
terms we a
ive at a subsequence. Also, if a sequence xn converges, and x denotes its
limit, then all but finitely many terms of the sequence are in the range [x− 0.1, x+
0.1], and these terms form a subsequence.
Lemma 2.1. Let a sequence xn converges to the limit x, and let xnk be a subsequence.
Then xnk → x as k →∞.
Proof. Fix ε > 0. Choose N such that
n ≥ N =⇒ |xn − x| < ε.
Then
k ≥ N =⇒ nk ≥ N =⇒ |xnk − x| < ε,
hence the sequence xnk tends to x. �
Theorem 2.4 (Bolzano–Weierstrass theorem). Every bounded sequence of real num-
ers has a convergent subsequence.
Proof. Let xn be a bounded sequence. Then {xn} ⊂ [a, b] for some a < b. We will
construct a subsequence xnk recursively, and first we have to say how the first term
is chosen.
At least one of the intervals [a, (a + b)/2] and [(a + b)/2, b] has infinitely many
terms of the sequence in it. Let a1 be and b1 be the endpoints of this interval, so
that if, for instance, [a, (a + b)/2] has infinitely many terms, we take a1 = a and
1 = (a+ b)/2. Then
(1) a ≤ a1 < b1 ≤ b,
(2) |b1 − a1| = 12 |b− a|, and
(3) there are infinitely many terms of xn in [a1, b1].
Let n1 be the integer such that xn1 is the first term of xn belonging to [a1, b1].
Further subintervals and terms in the sequence are chosen so that, at the p-th step,
we have already constructed numbers a1, a2, . . . , ap and b1, b2, . . . , bp (endpoints of
intervals) such that
(1) a ≤ a1 ≤ a2 ≤ · · · ≤ ap < bp ≤ · · · ≤ b2 ≤ b1 ≤ b,
(2) |bj − aj| = 2−j|b− a| for j = 1, 2, . . . , p, and
(3) there are infinitely many terms of xn in [ap, bp].
Indices n1 < n2 < · · · < np are chosen in such a way that xnj ∈ [aj, bj] for each
j = 1, 2, . . . , p. Let us now construct the next, (p + 1)-st, subinterval and term of
the subsequence.
22 WADIM ZUDILIN & BRAILEY SIMS
At least one of the intervals [ap, (ap + bp)/2] and [(ap + bp)/2, bp] has infinitely
many terms of the sequence in it; let ap+1 and bp+1 denote the endpoints of this
subinterval. Then
(1) a ≤ a1 ≤ a2 ≤ · · · ≤ ap ≤ ap+1 < bp+1 ≤ bp ≤ · · · ≤ b2 ≤ b1 ≤ b,
(2) |bj − aj| = 2−j|b− a| for j = 1, 2, . . . , p, p+ 1, and
(3) there are infinitely many xn in [ap+1, bp+1].
Choose np+1 to be the first integer larger than np and such that xnp+1 belongs to
[ap+1, bp+1].
With this construction in mind we obtain:
• a non-decreasing sequence ak bounded above by b,
• a non-increasing sequence bk bounded below by a, and
• a subsequence xnk with ak ≤ xnk ≤ bk for each k.
The monotone convergence theorem (Theorem 2.3) implies that the sequences ak
and bk converge; denote their limits A and B, respectively. We have
|A−B| ≤ |A−ak|+|ak−bk|+|bk−B| = |A−ak|+
1
2k
|b−a|+|bk−B| → 0 as k →∞,
hence A = B. We are now in a position to apply the squeeze theorem (Theorem 2.2)
to conclude that xnk → A as k →∞. Thus, xnk is a convergent subsequence of the
original sequence. �
Remark. Note that the property expressed in Theorem 2.4 serves in the definition of
compactness. Namely, a set S (of reals) is compact if every sequence chosen from S
has a subsequence which converges to an element of the set. With this in mind, we
have proved that any closed bounded segment [a, b] ⊂ R is compact.
MATH2330: ANALYSIS 23
A sequence xn is Cauchy if, given any ε > 0 there exists an N(ε) such that fo
m,n > N we have |xm − xn| < ε. In symbols we would write
∀ε > 0 ∃N(ε) such that m,n > N =⇒ |xm − xn| < ε.
Notice that this definition is in many ways similar to that for a converging sequence.
A Cauchy sequence has the property that eventually all the terms are a
itrarily
close to each other, but this does not necessarily mean that there is a limit to
the sequence. Imagine that we are working in the field of rational numbers, and
produce a sequence of rationals which is converging to
√
2. Such a sequence would
e a Cauchy sequence, but it would not have a limit within the field in which we are
operating, since
√
2 lies outside the rationals. Of course, if we were operating in the
field of real numbers it would have a limit. So Cauchy sequences are sequences which
are trying to have a limit, but may fail to do so because the thing which they want
to converge to is “missing”, like
√
2 is missing from the rationals. A set which has
“nothing missing”, so every Cauchy sequence actually has a limit, is called complete.
It is clear from the discussion above that the rationals are not complete. Our next
theorem is that the reals are.
Theorem 2.5 (Cauchy). A sequence xn of real numbers converges if and only if it
is Cauchy. That is, R is complete.
Note that the condition for being a Cauchy sequence applies to the terms of the
sequence only. Unlike the condition for convergence, we do not need to know what
the limit is.
Proof. “Only if ” (easier) direction. We show that, if a sequence converges, then it
is a Cauchy sequence.
Suppose that xn converges to xn → x, say. Given ε > 0, choose N such that
n ≥ N =⇒ |x− xn| < ε/2. Then for m and n larger than N we have
|xm − xn| ≤ |x− xm|+ |x− xn| < ε,
hence xn is a Cauchy sequence.
“If ” direction. Let xn be a Cauchy sequence.
For ε = 1 we take the co
esponding N = N(1) to ensure that |xm − xn| < 1 fo
all m,n > N . In particular, |xm − xN+1| < 1 for all m > N implying
|xm| = |xN+1 + (xm − xN+1)| ≤ |xN+1|+ 1
for those m. Therefore,
|xm| ≤ max{|x1|, |x2|, . . . , |xN |, |xN+1|+ 1}
for all indices m; that is, the sequence xn is bounded.
Since xn is bounded, the Bolzano–Weierstrass theorem ensures that there is a
convergent subsequence, xnk say. Let x denote the limit of this subsequence. We
claim that xn → x as n → ∞. Indeed, fix ε > 0 and choose N such that m,n ≥
N =⇒ |xm − xn| < ε/2. Since xnk → x, there is a k ∈ N such that nk ≥ N and
|x− xnk | < ε/2. Then
n ≥ N =⇒ |xn − x| ≤ |xn − xnk |+ |xnk − x| < ε;
24 WADIM ZUDILIN & BRAILEY SIMS
that is, the sequence xn converges to x. �
Exercise 2.20. Using the definition, verify whether the following sequences are Cauchy:
(a) an =
1
n
; (b) bn = (−1)n; and (c) cn = 1 +
1
2
+
1
3
+ · · ·+ 1
n
.
3. Limits of functions
Let S and T be sets. A function f : S → T is a rule which assigns to each element
x ∈ S an element f(x) ∈ T . The set S is called the domain of f and
f(S) := {f(s) : s ∈ S}
is called the range (or image) of f .
Note that a specification of the domain is part of the definition of a function. The
domain S of a function defined by a formula will usually be understood to be the
largest set for which the formula makes sense, for example dom(
√
x) = [0,∞).
3.1. Limits. Just as we did for sequences, we need to take the intuitive notion of
the limit of a function which we have all encountered in calculus courses, and make
that notion precise. The way in which this is done should not seem strange afte
our experience with sequences.
To say
lim
x→a
f(x) = L, (3.1)
means that as we force x to get close to a, the function value f(x) is forced close
to its limit L. As before with sequences, it is traditional to use ε as the measure
of closeness in the range. We need another small value to represent the degree of
closeness in the domain, and traditionally this role is filled by δ. Thus, we have the
definition (3.1) if
∀ε > 0 ∃δ(ε) > 0 such that 0 < |x− a| < δ =⇒ |f(x)− L| < ε.
Notice how the definition specifically excludes x = a. That is not to say the function
cannot have a value at x = a, just that its value there (if it has one) is i
elevant to
its limit at that point.
Exercise 3.1. Show that
lim
x→1
x2 − 1
x− 1
= 2.
Solution. Here we will have 0 < |x− 1| < δ (in particular, x 6= 1), so conside
|f(x)− L| =
∣∣∣∣x2 − 1x− 1 − 2
∣∣∣∣ = |(x+ 1)− 2| = |x− 1| < δ.
Now since we need to have |f(x)−L| < ε, we only need to ensure that δ ≤ ε. Thus,
if we choose δ = ε we win the “ε–δ game”. �
Exercise 3.2. Prove that
lim
x→2
(x2 + 1) = 5.
MATH2330: ANALYSIS 25
Solution. Here we will have 0 < |x− 2| < δ, so conside
|f(x)− L| = |(x2 + 1)− 5| = |x− 2| |x+ 2| < δ(δ + 4).
We therefore need to ensure that we can choose δ so that δ(δ + 4) < ε no matte
what ε > 0 we are given. One winning strategy is to solve this inequality and
conclude that 0 < δ < −2 +
√
4 + ε will do the trick. Another way is to say that
δ+ 4 is nearly equal to 4 and so is less than or equal to 5 (provided δ ≤ 1) and then
δ(δ + 4) ≤ 5δ < ε (provided δ < ε/5). Thus another winning strategy is to take
δ = min{1, ε/5}. �
In the examples above as well as in many further instances below, there is a
very good strategy to avoid unnecessary complications — the Ugly Thing Principle
(UTP). It says: seeing an ugly thing, give it a name. Introducing a new variable t,
namely, t = x−1 in Exercise 3.1 and t = x−2 in Exercise 3.2, we uniform the limits
lim
x→1
x2 − 1
x− 1
= lim
t→0
(1 + t)2 − 1
t
= lim
t→0
2t+ t2
t
= lim
t→0
(2 + t)
and
lim
x→2
(x2 + 1) = lim
t→0
(
(2 + t)2 + 1
)
= lim
t→0
(5 + 4t+ 4t2)
y having the same base t→ 0 in both of them. (The so different limits indeed look
very similar after the change!) In these examples x − 1 and x − 2 were the ugly
things.
Remark. You might notify that the powers of the new variable t are ordered not in
a customary way: first comes the t0- (constant) term, secondly the t1- (linear) term,
then t2-term, and so on (if you turn on your imagination). This is definitely not a
misprint and not accidental; we shall see the reasons behind this ordering very soon.
Perhaps the easiest way to show that a limit of a function does not exist is to
choose sequences of points from the function which have different limits. Conside
the function f(x) = sin(1/x). Of course, this function is not defined at x = 0, but
it might have a limit there. To see that this is not so, consider points x satisfying
1/x = (2n+ 1
2
)π where n ∈ N. It is easy to see that for these values of x the function
has value 1. Similarly, for those x satisfying 1/x = (2n − 1
2
)π where n ∈ N we see
the function has value...